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1

A simple model for bonding in this molecule is taking $\ce {Si}$ to adopt $\ce {sp^3}$ hybridisation, as suggested from the tetrahedral geometry of the molecule. Subsequently, we can refine our model by applying Bent's rule (see more about it here), which tells us that $\ce {p}$ orbital character concentrates towards more electronegative substituents. Since ...


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So here is the method I learned years ago without memorizing an awkward formula. The concept is to reduce the molecular formula to a hydrocarbon and compare it to the most saturated hydrocarbon bearing the same number of carbon atoms. Consider the infamous free-base, hydroxychloroquine, $\ce{C18H26ClN3O}$. The following atom replacements may be done in ...


3

The degree of unsaturation is defined as the index of hydrogen deficiency (IHD) that determines the total number of rings and π bonds. It means the removal of two hydrogen atoms from a molecule is equal to one added $\mathrm{DU}.$ $$\text{rings} + \pi~\text{bonds} = C - \frac{H}{2} - \frac{X}{2} + \frac{N}{2} + 1$$ If you add a halogen to a molecule, you ...


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Urea ($\ce{H2N-CO-NH2}$) is the quintessential amide. It is extremely soluble in water, and you could say it is a hydrogen donor and acceptor in water. However, if you replace one of the $\ce{-NH2}$ groups with an alkane or other organic group, solubility is reduced, depending on the size of the group. Amides generally have higher boiling points and melting ...


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The bond lengths given in your example tables are average bond lengths. That means, the actual bond length in given compound can be larger or smaller than the given value. Keep in mind that bond lengths are not just proportional to sizes of atoms involved making them. As pointed in the other answer, they are determine by other factors as well, which is a ...


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Bond length is not just proportional to atom size. Determine by other things too. One other determining factor of bond length is electronegativity. Bigger the difference in electronegativity the tighter the bond. I made up a table and stuff to illustrate this point but, as it so happens, the electronegativities didn't explain the data super well either. I ...


2

When you learn molecular orbital theory, you learn to throw the octet rule out the window in favor of identifying the molecular orbitals that form the actual bonds. Take a look at the seemingly innocuous case of carbon dioxide discussed in Chemistry LibreTexts (illustration from this reference): There are really two related ways in which the molecular ...


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The only intramolecular bonds in water are the O-H bonds. They are not modified during phase changes. The hydrogen bond is not an intramolecular bond. It is an intermolecular bond. These hydrogen bonds are modified during fusion or ebullition.


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OK, those who say magnetic fields have no effect can buy a relatively inexpensive Magnetizer (as I once did) and try performing a controlled experiment (as I attempted) with the rusting of iron. There is some apparent evidence that alignment of magnetic fields can promote radical activity. This UK study on the effect of a magnetizer in reducing the amount of ...


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As far as my sadly neglected physics knowledge goes It shouldn't. Oxidation is a chemical reaction process - electron exchange; and magnetism is spin dependent. Oxidation of metal will affect magnetism because it affects the physical structure of the material, but the inverse should not apply, as the energy drivers are order level lower for magnetization. ...


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Choices B and C are two different ways of saying the same thing. Say $E_{MO}(A)$ and $E_{MO}(AB)$ are the energies of electrons in bonding and anti-bonding orbitals, respectively, relative to the electrons in the original atomic orbitals from which the molecular orbitals are formed, or, to make matters simple, set the energy of electrons in the original ...


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When a test question is so vague that you can't figure out what answer is best, it's difficult to formulate a question about the answers to the question. The test question specifies bonding electrons equal to antibonding electrons. No energy statements. There needs to be one more answer: E. All of the above, although some answers are a little better than ...


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